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Which algorithm is faster O(N) or O(2N)?
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If an algorithm iterates over a list of numbers two times before returning an answer is the runtime O(2n) or O(n)? Does the runtime of an algorithm always lack a coefficient?
Big-O notation refers to the asymptotic "worst-case" complexity of an algorithm. Any constants are factored out of the analysis. Hence, from a theoretical perspective, O(2n) should always be represented as O(n). However, from the standpoint of practical implementation, if you can cut that down to one iteration over the list of numbers you will see some (small) increase in performance.
It may still be slower than an implementation that doesn't iterate twice, but that is still O(n), as the time complexity scales based only on the size of n.
Convention is that you ignore constant coefficients when reporting Big-O time.
So if an algorithm were O(n), O(2n), or O(3n) for example, you would report O(n).
Your suspicion is correct. You leave off the coefficient. See http://en.wikipedia.org/wiki/Big_O_notation.
From the example,
Now one may apply the second rule: $6x^4$ is a product of 6 and $x^4$
in which the first factor does not depend on x. Omitting this factor
results in the simplified form $x^4$.
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Prove that an algorithm has a lower bound
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Let's suppose we have a sorted list containing N elements. I have read in a textbook that an algorithm which determines if this list has duplicates must perform at least n-1comparisons (that means n-1 is a lower bound of the number of comparisons). I don't understand this because let's say the 1st and 2nd elements are duplicates, then the algorithm would simply return 'yes' after performing exactly one comparison. What do I get wrong? is there any simple proof of the n-1 lower bound?
When talking about complexity, you never consider only one input, otherwise there would always be some O(1) algorithm returning the expected answer.
An algorithm is supposed to work whatever the given input is (according to the specifications).
So for you, n-1 is the lower bound on the worst case complexity for any algorithm (based on comparison), i.e. for any algorithm you can find one input for which it needs at least n-1 comparisons.
You are probably confusing the two terms "lower bound" and "worst/best case".
The order of input determines the worst/best case, so, the lower bound or "big omega" can be said as "n-1" for the worst case and "1" for the best case.
But generally, the time complexity is determined for the worst cases.
This question has appeared in my algorithms class. Here's my thought:
I think the answer is no, an algorithm with worst-case time complexity of O(n) is not always faster than an algorithm with worst-case time complexity of O(n^2).
For example, suppose we have total-time functions S(n) = 99999999n and T(n) = n^2. Then clearly S(n) = O(n) and T(n) = O(n^2), but T(n) is faster than S(n) for all n < 99999999.
Is this reasoning valid? I'm slightly skeptical that, while this is a counterexample, it might be a counterexample to the wrong idea.
Thanks so much!
Big-O notation says nothing about the speed of an algorithm for any given input; it describes how the time increases with the number of elements. If your algorithm executes in constant time, but that time is 100 billion years, then it's certainly slower than many linear, quadratic and even exponential algorithms for large ranges of inputs.
But that's probably not really what the question is asking. The question is asking whether an algorithm A1 with worst-case complexity O(N) is always faster than an algorithm A2 with worst-case complexity O(N^2); and by faster it probably refers to the complexity itself. In which case you only need a counter-example, e.g.:
A1 has normal complexity O(log n) but worst-case complexity O(n^2).
A2 has normal complexity O(n) and worst-case complexity O(n).
In this example, A1 is normally faster (i.e. scales better) than A2 even though it has a greater worst-case complexity.
Since the question says Always it means it is enough to find only one counter example to prove that the answer is No.
Example for O(n^2) and O(n logn) but the same is true for O(n^2) and O(n)
One simple example can be a bubble sort where you keep comparing pairs until the array is sorted. Bubble sort is O(n^2).
If you use bubble sort on a sorted array, it will be faster than using other algorithms of time complexity O(nlogn).
You're talking about worst-case complexity here, and for some algorithms the worst case never happen in a practical application.
Saying that an algorithm runs faster than another means it run faster for all input data for all sizes of input. So the answer to your question is obviously no because the worst-case time complexity is not an accurate measure of the running time, it measures the order of growth of the number of operations in a worst case.
In practice, the running time depends of the implementation, and is not only about this number of operations. For example, one has to care about memory allocated, cache-efficiency, space/temporal locality. And obviously, one of the most important thing is the input data.
If you want examples of when the an algorithm runs faster than another while having a higher worst-case complexity, look at all the sorting algorithms and their running time depending of the input.
You are correct in every sense, that you provide a counter example to the statement. If it is for exam, then period, it should grant you full mark.
Yet for a better understanding about big-O notation and complexity stuff, I will share my own reasoning below. I also suggest you to always think the following graph when you are confused, especially the O(n) and O(n^2) line:
Big-O notation
My own reasoning when I first learnt computational complexity is that,
Big-O notation is saying for sufficient large size input, "sufficient" depends on the exact formula (Using the graph, n = 20 when compared O(n) & O(n^2) line), a higher order one will always be slower than lower order one
That means, for small input, there is no guarantee a higher order complexity algorithm will run slower than lower order one.
But Big-O notation tells you an information: When the input size keeping increasing, keep increasing....until a "sufficient" size, after that point, a higher order complexity algorithm will be always slower. And such a "sufficient" size is guaranteed to exist*.
Worst-time complexity
While Big-O notation provides a upper bound of the running time of an algorithm, depends on the structure of the input and the implementation of the algorithm, it may generally have a best complexity, average complexity and worst complexity.
The famous example is sorting algorithm: QuickSort vs MergeSort!
QuickSort, with a worst case of O(n^2)
MergeSort, with a worst case of O(n lg n)
However, Quick Sort is basically always faster than Merge Sort!
So, if your question is about Worst Case Complexity, quick sort & merge sort maybe the best counter example I can think of (Because both of them are common and famous)
Therefore, combine two parts, no matter from the point of view of input size, input structure, algorithm implementation, the answer to your question is NO.
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Why a programmer would prefer O(N^3) instead of O(N^2)
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Closed 7 years ago.
The order of growth of the running time of one algorithm is N; the order of growth of the running time of a second algorithm is N^2.
List TWO compelling reasons why a programmer would prefer to use the N^2 algorithm instead of the N one.
You cannot determine which algorithm is better base on order of growth. Order of growth just describes size of needed resources depend on input data. So there can be a situation where an program with worst complexity is faster, because input data isn't big enough.
For example Java implementation of sorting uses Insertion Sort ( O(N^2)) on small array (i.e. less that 29 elements) and counting sort ( O(N^2)) or Quicksort ( O(Nlog(N))) on big arrays.
Another example would be using Floyd-Warshall Algorithm ( O(N^3)) instead of Dijkstra Algortihm ( O(E*logV)) because it has much simpler implementation and doesn't need a priority queue (because we don't have its implementation in every environment)
If the actual running times are C0.N and C1.N² with C1.N < C0, it's better to use the "slow" algorithm.
Better space complexity.
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How is it different from asymptotic analysis? When do you use it, and why?
I've read some articles that seem to have been written well, like these:
http://www.ugrad.cs.ubc.ca/~cs320/2010W2/handouts/aa-nutshell.pdf
http://www.cs.princeton.edu/~fiebrink/423/AmortizedAnalysisExplained_Fiebrink.pdf
but I've still not fully understood these concepts.
So, can anyone please simplify it for me?
Amortized analysis doesn't naively multiply the number of invocations with the worst case for one invocation.
For example, for a dynamic array that doubles in size when needed, normal asymptotic analysis would only conclude that adding an item to it costs O(n), because it might need to grow and copy all elements to the new array. Amortized analysis takes into account that in order to have to grow, n/2 items must have been added without causing a grow since the previous grow, so adding an item really only takes O(1) (the cost of O(n) is amortized over n/2 actions).
Amortized analysis is not the same as an "average performance" - amortized analysis gives a hard guarantee on what the performance will do if you do so much actions.
There are a lot of answers to "what", but none to "why".
As everyone else has said, asymptotic analysis is about how the performance of a given operation scales to a large data set. Amortized analysis is about how the average of the performance of all of the operations on a large data set scales. Amortized analysis never gives worse bounds than asymptotic, and sometimes gives much better ones.
If you are concerned with the total running time of a longer job, the better bounds of amortized analysis are probably what you care about. Which is why scripting languages (for instance) are often happy to grow arrays and hash tables by some factor even though that is an expensive operation. (The growing can be a O(n) operation, but amortized is O(1) because you do it rarely.)
If you are doing real time programming (individual operations must complete in a predictable time), then the better bounds from amortized analysis don't matter. It doesn't matter if the operation on average was fast, if you failed to finish it in time to get back and adjust the bandsaw before it cut too far...
Which one matters in your case depends on exactly what your programming problem is.
Asymptotic analysis
This term refers to the analysis of algorithm performance under the assumption that the data the algorithm operates on (the input) is, in layman's terms, "large enough that making it larger will not change the conclusion". Although the exact size of the input does not need to be specified (we only need an upper bound), the data set itself has to be specified.
Note that so far we have only talked about the method of analysis; we have not specified exactly which quantity we are analyzing (time complexity? space complexity?), and neither have we specified which metric we are interested in (worst case? best case? average?).
In practice the term asymptotic analysis commonly refers to upper bound time complexity of an algorithm, i.e. the worst case performance measured by total running time, which is represented by the big-Oh notation (e.g. a sorting algorithm might be O(nlogn)).
Amortized analysis
This term refers to the analysis of algorithm performance based on a specific sequence of operations that targets the worst case scenario -- that is, amortized analysis does imply that the metric is worst case performance (although it still does not say which quantity is being measured). To perform this analysis, we need to specify the size of the input, but we do not need to make any assumptions about its form.
In layman's terms, amortized analysis is picking an arbitrary size for the input and then "playing through" the algorithm. Whenever a decision that depends on the input must be made, the worst path is taken¹. After the algorithm has run to completion we divide the calculated complexity by the size of the input to produce the final result.
¹note: To be precise, the worst path that is theoretically possible. If you have a vector that dynamically doubles in size each time its capacity is exhausted, "worst case" does not mean to assume that it will need to double upon every insertion because the insertions are processed as a sequence. We are allowed to (and indeed must) use known state to mathematically eliminate as many "even worse" cases as we can, even while the input remains unknown.
The most important difference
The critical difference between asymptotic and amortized analysis is that the former is dependent on the input itself, while the latter is dependent on the sequence of operations the algorithm will execute.
Therefore:
asymptotic analysis allows us to assert that the complexity of the algorithm when it is given a best/worst/average case input of size approaching N is bounded by some function F(N) -- where N is a variable
amortized analysis allows us to assert that the complexity of the algorithm when it is given an input of unknown characteristics but known size N is no worse than the value of a function F(N) -- where N is a known value
The answer to this is succinctly defined by the first sentence of the Amortized Analysis chapter in the book - Introduction to Algorithms:
In an amortized analysis, the time required to perform a sequence of
data-structure operations is averaged over all the operations
performed.
We represent the complexity of a program's growth by Asymptotic analysis - which is bounding the program's growth by a function and defining the worst, best or average case of that.
But this can be misleading in cases where there is just one case where the program's complexity reaches a peak, but in general, the program doesn't take much computation.
Hence, it makes more sense to average the cost over a sequence of operations, even though a single operation might be expensive. This is Amortized Analysis!
Amortized Analysis is an alternate to Asymptotic technique used to calculate complexity. It helps us calculating a more true complexity in terms of practicality, so as to compare and decide between two or more algorithms.
The best reference I've found so far for understanding the amortized analysis of algorithms, is in the book Introduction to Algorithms, third edition, chapter 17: "Amortized Analysis". It's all there, explained much better than what can be found in a Stack Overflow post. You'll find the book in the library of any decent University.
Regular asymptotic analysis looks at the performance of an individual operation asymptotically, as a function of the size of the problem. The O() notation is what indicates an asymptotic analysis.
Amortized analysis (which is also an asymptotic analysis) looks at the total performance of multiple operations on a shared datastructure.
The difference is, amortized analysis typically proves that the total computation required for M operations has a better performance guarantee than M times the worst case for the individual operation.
For example, an individual operation on a splay tree of size N can take up to O(N) time. However, a sequence of M operations on a tree of size N is bounded by O( M(1+log N) + N log N ) time, which is roughly O(log N) per operation. However, note that an amortized analysis is much stricter than an "average-case" analysis: it proves that any possible sequence of operations will satisfy its asymptotic worst case.
Amortised analysis deals with the total cost over a number of runs of the routine, and the benefits that can be gained therein. For example searching an unsorted array of n items for a single match may take up to n comparisons and hence is o(n) complexity. However, if we know the same array is going to be searched for m items, repeating the total task would then have complexity O(m*n). However, if we sort the array in advance, the cost is O(n log(n)), and successive searches take only O(log(n)) for a sorted array. Thus the total amortised cost for m elements taking this approach is O(n*log(n) + m*log(n)). If m >= n, this equates to O(n log(n)) by pre-sorting compared to O(n^2) for not sorting. Thus the amortised cost is cheaper.
Put simply, by spending a bit extra early on, we can save a lot later.
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Quicksort has a worst-case performance of O(n2), but is still used widely in practice anyway. Why is this?
You shouldn't center only on worst case and only on time complexity. It's more about average than worst, and it's about time and space.
Quicksort:
has average time complexity of Θ(n log n);
can be implemented with space complexity of Θ(log n);
Also have in account that big O notation doesn't take in account any constants, but in practice it does make difference if the algorithm is few times faster. Θ(n log n) means, that algorithm executes in K n log(n), where K is constant. Quicksort is the comparison-sort algorithm with the lowest K.
Average asymptotic order of QuickSort is O(nlogn) and it's usually more efficient than heapsort due to smaller constants (tighter loops). In fact, there is a theoretical linear time median selection algorithm that you can use to always find the best pivot, thus resulting a worst case O(nlogn). However, the normal QuickSort is usually faster than this theoretical one.
To make it more sensible, consider the probability that QuickSort will finish in O(n2). It's just 1/n! which means it'll almost never encounter that bad case.
Interestingly, quicksort performs more comparisons on average than mergesort - 1.44 n lg n (expected) for quicksort versus n lg n for mergesort. If all that mattered were comparisons, mergesort would be strongly preferable to quicksort.
The reason that quicksort is fast is that it has many other desirable properties that work extremely well on modern hardware. For example, quicksort requires no dynamic allocations. It can work in-place on the original array, using only O(log n) stack space (worst-case if implemented correctly) to store the stack frames necessary for recursion. Although mergesort can be made to do this, doing so usually comes at a huge performance penalty during the merge step. Other sorting algorithms like heapsort also have this property.
Additionally, quicksort has excellent locality of reference. The partitioning step, if done using Hoare's in-place partitioning algorithm, is essentially two linear scans performed inward from both ends of the array. This means that quicksort will have a very small number of cache misses, which on modern architectures is critical for performance. Heapsort, on the other hand, doesn't have very good locality (it jumps around all over an array), though most mergesort implementations have reasonably locality.
Quicksort is also very parallelizable. Once the initial partitioning step has occurred to split the array into smaller and greater regions, those two parts can be sorted independently of one another. Many sorting algorithms can be parallelized, including mergesort, but the performance of parallel quicksort tends to be better than other parallel algorithms for the above reason. Heapsort, on the other hand, does not.
The only issue with quicksort is the possibility that it degrades to O(n2), which on large data sets can be very serious. One way to avoid this is to have the algorithm introspect on itself and switch to one of the slower but more dependable algorithms in the case where it degenerates. This algorithm, called introsort, is a great hybrid sorting algorithm that gets many of the benefits of quicksort without the pathological case.
In summary:
Quicksort is in-place except for the stack frames used in the recursion, which take O(log n) space.
Quicksort has good locality of reference.
Quicksort is easily parallelized.
This accounts for why quicksort tends to outperform sorting algorithms that on paper might be better.
Hope this helps!
Because on average it's the fastest comparison sort (in terms of elapsed time).
Because, in the general case, it's one of the fastest sorting algorithms.
In addition to being the fastest though, some of it's bad case scenarios can be avoided by shuffling the array before sorting it. As for it's weakness with small data sets, obviously isn't as big a problem since the datasets are small and the sort time is probably small regardless.
As an example, I wrote a python function for QuickSort and bubble sorts. The bubble sort takes ~20 seconds to sort 10,000 records, 11 seconds for 7500, and 5 for 5000. The quicksort does all these sorts in around 0.15 seconds!
It might be worth pointing out that C does have the library function qsort(), but there's no requirement that it be implemented using an actual QuickSort, that is up to the compiler vendor.
Bcs this is one of Algorithm with work well on large data set with O(NlogN) complexity. This is also in place algorithm which take constant space. By selecting pivot element wisely we can avoid worse case of Quick sort and will perform in O(NlogN) always even on sorted array.